# Dual Spin satellite control using momentum exchange

When the stability of dual spin satellite with rotor along z axis an additional equation for the relative motion between the rotor and satellite is added to Euler's equation. Similarly on introducing a reaction wheel a relative motion equation is added to the system. If I introduce a reaction wheel to dual spin satellite, does my dynamics equation remain the same for a dual spin or does now my satellite becomes three axis spin stabilized?

edit: eg. If I have a rotor along $Z$ then I use modified Euler equations with relative dynamics of rotor given by $M_r = I_r \cdot \dot \omega$, and the moment equation given by $I_i \cdot \dot \omega_i + S_{ij} + I_r \cdot \omega_{rk} = M$, where $i,j,k$ represents the corresponding axes, and $S_{ij}$ is the difference of inertia. At this point I want to introduce 3 reaction/inertia wheels aligned with each axes. The required torque is computed using the standard control equations, which goes into the above mentioned dynamics equation. But at this point does my dynamics equation changes with the introduction of new rotors along each axes or does it stays the same as before.

• Could you please copy edit for clarity? I can't speak for others, but I find your question rather confusing and some sentences seem to be missing parts of them. Thanks! – TildalWave Jun 22 '15 at 15:32
• does it make sense now? – Astroynamicist Jun 22 '15 at 17:09
• Not that much, actually the first sentence lacks a verb (at least). And in general I think it is still not clear. – tuspazio Jul 2 '15 at 16:23

Given that:

• Euler equations are general, there exist no special or modified forms. Simply some terms are equal to zero if some conditions are satisfied
• I do not quite understand what $S_{ij}$ is in your expression

Anyway, if the centre of mass of the body coincides with the origin of the body frame, the total angular momentum of the satellite can be written as $$\underline{H} = J\underline{\omega} + \underline{h}$$

Where $J$ is the inertia tensor, $\underline{\omega}$ is the angular velocity vector, $\underline{h}$ is the total angular momentum of all the rotating internal equipments. The second law of motion states that the derivative of the total angular momentum equals the sum of the external torques

$$\frac{D\underline{H}}{Dt}=\underline{\dot H}+\underline{\omega}\times\underline{H}=\underline{M}$$

Neglecting the variation of the inertia tensor, the previous can be written as $$J\underline{\dot \omega}+\underline\omega\times J\underline\omega + \underline {\dot{h}} +\underline\omega\times\underline h=\underline M$$ If you have 3 rotors each aligned with one of the satellite axes, then $\underline h$ is $$\underline h = \begin{bmatrix} J_{r_1}\omega_{r_1} & J_{r_2}\omega_{r_2} & J_{r_3}\omega_{r_3} \end{bmatrix}^\rm T$$ You can name the vector of control torques (coming from some control technique as you mentioned) as $$\underline{M}_c = -\underline {\dot{h}} -\underline\omega\times\underline h$$

and compute the needed torque each wheel needs to provide

$$\underline {\dot{h}} = -\underline{M}_c -\underline\omega\times\underline h$$

Hope this helps